Let $E$ be an elliptic curve over a field of characteristic $p$, and let $n$ be an integer coprime to $p$. Then $E[n]$, the kernel of multiplication by $n$ on $E$, is (etale-locally) isomorphic to $(\mathbb{Z}/n)^2$. A level-$n$-structure on $E$ is a specification of such an isomorphism $(\mathbb{Z}/n)^2\to E[n]$.

Now $GL_2(\mathbb{Z}/n)$ acts on the set of level structures by pre-composition, and even more, the set of level-$n$-structures on $E$ is a $GL_2(\mathbb{Z}/n)$-torsor over the point on the moduli stack of elliptic curves determined by $E$ (I mean here the stacky point $B Aut(E)$).

The group $G:=Aut(E)$ acts on the set of level structures on $E$ by post-composition. This determines a morphism $\rho: G \to GL_2(Z/n)$. Is this morphism explicitly known in general?

I can describe this representation in the specific cases of $p=2$ and $n=3$, or $p=3$ and $n=2$, by writing down explicit equations for the curve and knowing explicitly in terms of those equations what do choices of points of the given order look like. But these are presumably more complicated cases in the sense that the structure of $G$ can be more complicated at $p=2$ or $3$.

However, if the characteristic is not $2$ or $3$, $G$ is isomorphic to $\mu_k$, where $k$ is $2,4$ or $6$, depending on the $j$-invariant. One could write down representations of these $\mu_k$ in $GL_2(\mathbb{Z})$ that descend to representations in $GL_2(\mathbb{Z}/n)$ for all $n$. (One condition we probably need is that $-1\in G$ maps to the identity matrix.) For example,
$\left(\begin{smallmatrix}
-1 & -1 \\\
1 & 0
\end{smallmatrix}\right)$
has order $3$, and
$\pm\left(\begin{smallmatrix}
0 & 1 \\\
1 & 0
\end{smallmatrix}\right)$
has order 2. How can I know if (or when) these matrices determine the representations I am looking for?

Shouldn't the level-$n$ structures agree with Cartier duality, making it $SL_2$?
–
Will SawinFeb 4 '12 at 20:05

The 2/4/6 automorphisms extend to the elliptic curve over $\mathbb Z$, and therefore to a corresponding curve over $\mathbb Q$ or even $\mathbb C$. You can therefore look at pictures of the appropriately symmetric lattices and see how the automorphisms act.
–
Will SawinFeb 4 '12 at 20:08

That's how I got the matrices above, the question is whether the argument really goes through for the general case.
–
Vesna StojanoskaFeb 5 '12 at 0:29

This might work: Compute the $j$ invariant mod $p$, it has to be $0$ or $1728$ or there are no extra automorphsims. Choose a standard curve for that $j$-invariant with good reduction mod $p$, show they're isomorphic mod $p$. If it has only standard automorphisms, then it certainly has $X\to -x$, so that has to be the one it has.
–
Will SawinFeb 5 '12 at 4:41